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I am currently reading Linear Algebra Done right and was having trouble with the proof of the Real Spectral Theorem.

Statement:

7.13 Real Spectral Theorem: Suppose that V is a real inner-product space and T ∈ L(V). Then V has an orthonormal basis consisting of eigenvectors of T if and only if T is self-adjoint.

The theorem he is refering to in the proof is:

7.12 Lemma: Suppose T ∈ L(V) is self-adjoint. Then T has an eigenvalue.

I am only interested in the proof from one side. The proof is a bit long so I will just post the picture:

enter image description here

enter image description here

It isint fully clicking to me. He defined an artificial operator himself and is saying this must have an orthogonal basis even though he never showed it for matrices greater than dimension 1. Dimension 1 is such an unrealistic case for matrices. Can anyone clear out his thinking or approach for me?

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    $\begingroup$ Please don't post pictures of text: they are not searchable, they sometimes have bad interactions with different interfaces, and finally, they often provide no support for those who use accessibility tools. $\endgroup$ – Arturo Magidin Aug 3 at 4:10
  • $\begingroup$ I understand but as I stated the proof was a bit long. And pictures can be made searchable very easily. Anyways care to answer? $\endgroup$ – Rahul Deora Aug 3 at 6:32
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    $\begingroup$ So, in short, you know but you don’t care, your convenience being of paramount importance. Gotcha. No, pictures are not searchable on this site. If someone searches for the text in your picture using the native search, it will not show up. As to answering the question, I know the answer, but it is a bit long and I don’t feel like typing it. $\endgroup$ – Arturo Magidin Aug 3 at 16:55
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The proof is by induction on the dimension of $V$; as with all proofs by induction, that means that we need to explicitly show that the statement is true for some base case(s) (in this case, when the dimension of $V$ is $1$), and that if the statement is true up to some dimension $n$ then it remains true in dimension $n+1$.

The approach is to take a vector space $V$ of dimension $n+1$ and breaking it up into two pieces, namely the subspace $U$ spanned by an eigenvector of $T$ and the subspace $U^\perp$ that is orthogonal to $U$. If $\alpha$ is an orthonormal basis of $U$ and $\beta$ is an orthonormal basis for $U^\perp$, then $\alpha \cup \beta$ is an orthonormal basis for $V$, so all we need to do is find $\alpha$ and $\beta$, each consisting of eigenvectors of $T$.

Finding $\alpha$ is easy, because $U$ is 1-dimensional and spanned by an eigenvector of $T$; just take any vector in $U$ and scale it to have norm $1$.

To find $\beta$ we'd like to apply the induction hypothesis. We do have $\dim(U^\perp) = n < \dim(V) = n+1$, which is good: If we have a self-adjoint operator from $U^\perp$ to $U^\perp$ then the induction hypothesis will give us the basis for $U^\perp$ that we're looking for. The operator we'd like to use is $T$, but $T$ is an operator from $V$ to $V$, not from $U^\perp$ to $U^\perp$. It would be nice, though, if we could think of $T$ as an operator from $U^\perp$ to $U^\perp$. For that reason we define $S : U^\perp \to U^\perp$ to do the same thing as $T$: for all $v \in U^\perp$, $S(v) = T(v)$. There's a little checking to do to make sure this makes sense (specifically, that if $v \in U^\perp$ then $T(v) \in U^\perp$ too), and that $S$ is self-adjoint.

Once those steps are done, we've now got a space ($U^\perp$) of dimension strictly less than the dimension of $V$, and a self-adjoint operator on that space. By induction hypothesis there is an orthonormal basis, call it $\beta$, for $U^\perp$ consisting of eigenvectors of $S$. But $S$ does the same thing as $T$, so the vectors in $\beta$ are also eigenvectors of $T$, which is what we wanted.

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